1. Oblicz ile moli stanowi:
a) 1 kg żelaza b) 11,2 dm3 H2( warunki normalne)
c) 0,5 gram tlenku magnezu.
1000g Fe------xmoli
56g-------1mol
xmoli=17,86mola
b)11,2dm3H2-----xmoli
22,4dm3H2----1mol
xmoli=0,5mola
c) 0,5gMgO------xmoli
40gMgO---------1mol
xmoli=0,0125mola
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1000g Fe------xmoli
56g-------1mol
xmoli=17,86mola
b)11,2dm3H2-----xmoli
22,4dm3H2----1mol
xmoli=0,5mola
c) 0,5gMgO------xmoli
40gMgO---------1mol
xmoli=0,0125mola